WSSS 2021 Prelim P2 ANS

SECTIONA(50MARKS) 1a) 1b) 2a) Productofforce[1]andperpendiculardistance[1]topoint 2bi) W=mgW=(0.080)(10)W=0.80N[1] 2bii) [1]𝐹 × 0.75 = 0.8 × 0.25[1]𝐹 = 0.27𝑁 2c) PerpendiculardistancefrompivotofFdecreases/islessthan0.75m[1]ORPerpendiculardistancefrompivotofWincreases/ismorethan0.75m,hence

increased/greaterforceF(forgreatermoment)[1] 3a) [1]𝑃 = 𝐸𝑡 × 10085 = 75×10×1008.5 × 10085 [1]= 1.04 × 10 4 𝑊 3b) [1] 12 × 75 ×𝑣 2 = 75 × 10 × 41[1]𝑣 = 28.6𝑚/𝑠 3ci) GravitationalPotentialEnergy+KineticEnergy[1]ElasticPotentialEnergy(+heatenergyinrope)[1]→ 3cii) GaininG.P.E=LossinK.EGaininG.P.E=75 × 10 × 73GainInG.P.E=54800J[1] 4a) 𝑃 = ℎρ𝑔 +𝑃𝑎𝑡𝑚 P=(0.400)(1000)(10)+100000[1] [1] P=104000Pa 4bi) - Moleculesmoveslower[1]- Collidelessfrequently[1]withwall- Forceonwalldecreases[1] 4bii) Waterlevelonleftsideishigherthanlevelontheright[1] 5a) 𝑄 = 𝑚𝑐∆θQ=(0.2)(4600)(65-10)[1]Q=50600J[1] 5b) 𝑚× 2.2 × 10 6 + [𝑚× 4200 × (100 − 65)] = 50600[1] [1] m=0.0216kg[1] 5c) Steamcontainsmore(latent)heatenergy[1]thanwaterSpecificlatentheatofvaporisationismuchgreaterthan[1]specificheatcapacityofwater 6a) Pointatwhichraysconvergethatwereinitiallyparalleltoprincipalaxisoflens[1]

6bi)6bii) 6c) LessthanfocallengthORbetweenpointFandlens[1] 6d) Converging 7ai) 7aii) Electronsareattractedandmovetowardstherod[1]andclusterontheleftsideofKsinceunlikechargesattract,leavingbehindinducedpositivechargesontherightsideofL.[1] 7bi) Electronsmoveandredistributethemselves[1] 7bii) Neutralised[1] 7c) Placeitoverintenseflame[1]

8a) Workdonetodriveaunitchargeacrossanentirecircuit[1] 8bi) ε = 𝐸𝑄 = 1.6×10 5 1.8×10 4 =8.9V[1]ε 8bii) 𝐼 = 𝑄𝑡 = 1.80×10 5 1.3×10 5 I=0.14A[1] 8ci) [1]𝐸 = 1545 × 1.1 × 10 5 [1]𝐸 = 3.7 × 10 4 𝐽 8cii) Extraresistanceincircuitfromwires,ammeters,cells[1] SECTIONB(30MARKS) 9a) - Equalinmagnitudeandoppositeindirection[1]- Onescapinggasfromrocket[1] 9b) Thevelocityoftherocketincreases/acceleratesfrom0to0.80s.Thisshowsthattheresultantforceisupwards,henceliftispresentandgreaterthanweight.[1]After0.8s,accelerationisnegative[1].Hence,resultantforceisdownward,duetoweightonlyandthusthereisnolift.[1] 9c) - Sincerocket2hashigheracceleration,itislikelytohaveasmallermass.- andifliftforceisthesame,then and𝐹𝑛𝑒𝑡 = 𝐿𝑖𝑓𝑡 𝑓𝑜𝑟𝑐𝑒 − 𝑤𝑒𝑖𝑔ℎ𝑡 𝐹𝑛𝑒𝑡 accelerationwillbelarger.[1] 9di) 1.40s[1] 9dii) [1]ℎ𝑚𝑎𝑥 = 12 × 10 × (2.4 − 1.4)[1]ℎ𝑚𝑎𝑥 = 5.0𝑚 10ai) (imaginary)lineonawavethatjoinsallpeaksofawave[1] 10aii) Transferofenergyfromonepointtoanotherthroughvibrations[1]withouttransferringmatter.[1]Thedirectionofwavemotionisperpendiculartothedirectionofvibrationofwatermolecules.[1] 10bi) Difference:Differentamplitude[1]Similarity:Samefrequency[1]

10bii) 1. [1]𝑓 = 1𝑇 = 1300×10 −3 f=3.33Hz[1] 2. 𝑣 = 𝑓λ [1]0.20 = 3.33λ[1]λ = 0.0601𝑚 11ai) Themagneticfieldofthemagnetsandmagneticfieldofcurrentinwireinteractandproduceaforce.[1] 11aii) Fleming’slefthandrule.Directionofforcebythumb,directionofmagneticfieldbyindexfingeranddirectionofcurrentbysecond(middle)fingerareperpendiculartooneanother.[3] 11b) - ThereisnoforceForitdisa